I need some help on following question. Any help is greatly appreciated.

Let $p$ be a sigma finite measure on measurable space $( [0,1], M )$ where $M$ is the sigma algebra of Lebesgue measurable subsets of $[0,1]$. For $x$ in $[0,1]$, define $F(x) = p ([0,x])$.

Show that $p$ is absolutely continuous with respect to Lebesgue measure $m$ on $[0,1]$ iff $F$ is an absolutely continuous function on $[0,1]$ satisfying $F(0) = 0$.

  • $\begingroup$ Do you have any thoughts and can share those? Regards $\endgroup$ – Amzoti Aug 8 '13 at 17:19
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    $\begingroup$ I actually worked on it but not sure about this. Is it correct that F(b) - F(a) = p([a,b]) for 0<a<b<1 regarding measure p? I am not sure what p does..Is it an aritrary measure or it is just same as lebesgue measure on [0,x]? $\endgroup$ – Kushan Aug 8 '13 at 17:25
  • $\begingroup$ Thanks for sharing your attempt in the comment. You might like to edit your question with the details of what you've tried so far. It makes it a lot easier for someone to know what part of the problem you're having trouble with, and also lets us know you're not just here to get easy answers to a homework problem (it's unfortunate but it happens :-/ ). $\endgroup$ – Dan Rust Aug 8 '13 at 17:33
  • $\begingroup$ Thanks for the comments. I am working on prelim questions...I did work it out, but only part I was not sure about was what I have mentioned earlier since I used it.I guess I would have mentioned it with the question..Sorry for that. $\endgroup$ – Kushan Aug 8 '13 at 17:38

It seems, from your comments, that the part you're asking about is the following statement, which I've made slightly more precise than you wrote.

Claim: For every $0<a<b<1$, we have $F(b)-F(a) = p\big( (a,b] \big)$.

Proof: Note that $[0,b] = [0,a] \cup (a,b]$ and that this is a disjoint union. Since $p$ is a measure, we have $p\big( [0,b] \big) = p\big( [0,a] \big) + p\big( (a,b] \big)$. So by the definition of $F$, we have $F(b) = F(a) + p\big( (a,b] \big)$ as claimed.

(Note that it is entirely possible for $p\big( (a,b] \big) \ne p\big( [a,b] \big)$ - for example, $p$ could equal $1$ on every set containing $a$ and $0$ on all other sets. Of course that doesn't happen for absolutely continuous measures, since the Lebesgue measure of $\{a\}$ equals $0$; but this question requires thinking about all measures.)


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